An object with a mass of #9 kg# is on a plane with an incline of # - pi/4 #. If it takes #15 N# to start pushing the object down the plane and #5 N# to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Nov 20, 2017

Answer:

#\mu_s = (F_{app}^s/(mg\cos\theta) + \tan\theta) = 1.24#
#\mu_k = (F_{app}^k/(mg\cos\theta) + \tan\theta) = 1.08#

Explanation:

Consider a body of mass #m# on an incline with an inclination angle of #\theta#. To this a force is applied without disturbing the equilibrium. The forces acting on the object are
#vec W # : Weight of the object acting vertically downward;
#vec F_{app} # : Force applied parallel to the inclined plane,
#vecF_f# : Frictional force acting parallel to the plane of the incline,
#vec N # : Normal force acting perpendicular to the plane of the incline,

Consider a coordinate system with its X-axis parallel to the plane downward and Y-axis perpendicular to the plane upward.

Resolve the weight into components that are parallel and perpendicular to the inclined plane

#vec W_{||} = -mg\sin\theta; \qquad vec W_{\_|_} = -mg\cos\theta#

Equilibrium: #\qquad vec W + vec F_{app} + vec N + vec F_f = vec 0#

(A) Perpendicular Component: #\qquad vec N + vec W_{\_|_} = vec 0#

#N - W_{\_|_} = 0; \qquad N = mg\cos\theta;#

(B) Parallel Component: #\qquad vec W_{||} + F_{app} + vec F_f = vec 0#

#W_{||} + F_{app} - F_f = 0; \qquad F_f = F_{app} + W_{||} # ...... (2)

Static Equilibrium: In this case the frictional force is due to static friction.

#F_f = \mu_sN = \mu_s.mg\cos\theta; \qquad F_{app} = F_{app}^s = 15# #N;#

Expanding (2), #\mu_s.mg\cos\theta = F_{app}^s + mg\sin\theta#
#\mu_s = (F_{app}^s/(mg\cos\theta) + \tan\theta)# ...... (3)

Dynamic Equilibrium: In this case the frictional force is due to kinetic friction.

#F_f = \mu_kN = \mu_k.mg\cos\theta; \qquad F_{app} = F_{app}^k = 5# #N;#

Expanding (2), #\mu_k.mg\cos\theta = F_{app}^k + mg\sin\theta#
#\mu_k = (F_{app}^k/(mg\cos\theta) + \tan\theta)# ...... (4)

Calculate:
#\sin\theta = \cos\theta = 1/\sqrt{2}; \qquad \tan\theta = 1; \qquad mg\cos\theta = 62.37# #N#
#mu_s = 1.24; \qquad \mu_k = 1.08#